SA17: Shear Influence Line

[@k.w.9242]

Best tutorial ever. Well presented. If I were rich I would fund people like you. Thanks for making student's lives easier. God bless!

[@tauseefraza3094]

Very high thinking May Allah make you capable of this

[@a-02sagarjadhav53]

I too 👍10000000000.....K👍

[@abhishekmina3996]

You deserve a gold medal ...... Just cleared my concept in just 15 minutes , which I was facing from last two year

[@INDIA_PRIDE_I]

This channel is a gem.

[@thobanigoba2024]

I’m watching this from my structural mechanics class and in 15 minutes I assure you I’ve learned more than 75% of the people listening to the lecturer for the passed 45 minutes. Quality video👌🏽👌🏽👌🏽

[@musica3955]

ahahahahha Savage

[@87455451jose]

Amazing, I've been struggling with this for weeks, and I got it all cleared in 15 min. Cant wait to see SA18 moment influence lines. Cheers!

[@FaizYah]

God bless you Dr. ! Thank you for such an easy-to-understand explanation!

[@ahmadshahi4421]

This was the greatest training videos i ever seen in youtube

[@abeal-qudah1619]

Thank you. I was to understand influence lines from you video that 5 years of education could not do for me. Thank you again!

[@AhmetOzdemir-lk1bl]

)))

[@omohosagie4772]

Demystifying....You are truly a Doctor of Structures!

[@azhysurkew]

idk how to thank u,, this influence line has been a nightmare for me but u make it very simple and easy ,, much respect

[@patrickwoolery6000]

Thank you so much for these videos. My SA professor spent like 5 minutes on influence lines and thought we'd all understand. I don't know why he didn't just play your videos for the class because these teach the material so much better. Again, thank you.

[@DrStructure]

Thank you for the feedback.

[@لاتحزن-ف9ك]

@@DrStructure am from Iraq and very difficult to understand English language

[@ibrahimacar8262]

Whoever found this is out of this world bro, It's epic. You don't need functions to solve anymore.

[@musichub6752]

best explanation i have ever experience in my student life . Thanks alot

[@asaddavar43]

This is the best youtube video for influence line of beam so far.... thanks from bd..

[@tabzz03]

I can't thank you enough. I was breaking my head from past three days on drawing ild. I saw numerous videos on yt but thy didn't help. Thank you so much for making us understand in such simple way. God bless 🙏

[@reginavillalobos1301]

OMG years of trying to understand this concept and I just finally did! Thank you so much!

[@DrStructure]

You’re welcome.

[@lebogangtsietsi3608]

I just love you Dr Structure....

[@wongsathornphatcharaphisut3020]

This video is better than my in class lecturer. I can understand all of a chapter just in quarter an hour. Thank you so much Dr.

[@Kamikorosu0416]

Unbelievably amazing! Very clear and understandable. Cheers from the Philippines

[@_jojo11]

this is a godsend a few days before structure exams thank you

[@prakashpandey4907]

You have mind blowing teaching style....

[@rahulmalpekar4631]

Thankyou so much @Dr. Structure for providing this wonderful video. It was really concept clearing.

[@bishalnayak6162]

God bless you to whoever made this video.

[@coolbro1502]

You deserve to be respect sir

[@moisestorrespe]

Excelente video!!! It helps me a lot in my master's advanced bridge design class

[@ieuhfuffffff]

I'm Vietnamese, I'm not good at in English, but I can understand all of things you explain. At my University, I cant understand it. Thank you very much, you are a master in explain.

[@DrStructure]

Thanks for your comment and feedback.

[@tuze5474]

OMG you helped me a lot, now I understand how it works! Marvelous video!

[@1carlos83]

Thank you for this, the graphics are the best I have seen for anyone lecturing structures. You should do a P.E. course.

[@عمارسامي-ز7ص]

الله يباركلك يا هندسه❤

[@johnkim4026]

Dude i have a final tomorrow and this saved my life. U are amazing

[@junielvillas8820]

for 4:10, is it possible to be drawn the other way around, like upward force at the left roller and downward at right?

[@DrStructure]

By definition, shear in a beam segment is considered positive when the force is acting downward at the right end of the segment, and upward at the left end of the segment. That is what you see @4:10, positive shear in both segments. If we define positive shear the other way around, then yes, we would have to switch the direction of the arrows.

[@eminashindahouse]

Dear Dr. Why at 3:04 , would you describe the shear at C decreases linearly when the load moving from A to C? Isn’t it increases linearly? My understanding is that the negative sign only indicates the right hand side of shear with respect to the load at current time.

[@DrStructure]

If the value of shear is increasing in the positive direction, we say shear is increasing. On the other hand, if the value of shear is moving in the opposite direction, we say it is decreasing. That is, if shear goes from 0 to, say, -100, we say shear is decreasing, since -100 is smaller than 0.

[@eminashindahouse]

Dr. Structure understood. Many thanks 🙏🏻🙏🏻

[@DrStructure]

You're welcome.

[@madisonstafford6619]

You actually just saved my grade.

[@noorhanalkhatib4341]

I love this channel .. amazing

[@kaylal3875]

The best video out there! Thank you so much!

[@pronoybhat]

What a way to learn Structure.. Great work. Thanks !

[@UnknownUser12]

Thank you for your videos and putting so much effort in them. Regards from Aus

[@love-and-faith]

You deserve gold you have to teach other teachers how to teach 😮😮😮😮

[@KonjoTube]

u save my life i love u Dr.

[@franciscovasquezramirez5705]

Thank Dr Structure; Congrats from Bolivia

[@ahamadromen]

A wonderful work. Thank you for your v.well simple explanation.

[@smboston12]

this is so great, makes it so easy to understand. Thank you

[@manikantagodi3239]

is that that much hard? my reaction when I heard this topic in the class...but after this video is that that much easy?...overturned my reaction...thank you so much..you just nailed the topic...

[@veersundarr7821]

Very nice explanation!

[@fatimamancuso]

On the example with shear @ E, rollers B & C on either side, (@7:54 )with a hinge at D. What if B & D were switched to: hinge at B and roller at D instead? The shear with a hinge next to it on the left, roller to the right, and roller to the left of the hinge? I did not see an example with a hinge and then a roller next to the IL of question, only saw rollers side by side or roller / hinge combo. Does it act like the example with two rollers where the bar cannot move and the IL goes to the hinge and stays there? Thank you! Hope that makes sense.

[@DrStructure]

Thanks for the question. Yes, in such a case the left segment of the beam from A to B (the new location of the real hinge) remains flat. E moves downward to the left of the fictitious roller at E, causing BE to rotate clockwise. And, E moves upward to the right of the fictitious roller making EC to turn clockwise as well.

[@fatimamancuso]

Makes sense!!! I appreciate it!

[@rmrafsanzanilabib6881]

You explained IL so well. thank you for making our study easier

[@osuwatchdog7154]

thanks for saving my life

[@gopalkumar-mh7zx]

thanks from India ..for make understanding concept very clealy

[@rubencasinillojr9852]

Thank you! Great explanation ever.

[@ahmadabdelgalil5894]

Thanks 💖💖💖

[@yadrzgar9980]

A structural analysis professor at my uni who has a PhD & 12 years experience could bot explain this half as good as this video! Thank you 😭😍

[@omarpasha2968]

I believe it! I once took my junior College electrical problems to a university electrical engineering tutor. And, the problem was too complicated for him!

[@ibrahimyasser4719]

Great thanks and great respect

[@Mm-hw9yb]

Thanks a lot Dr. Structure!

[@gnidnoeled786]

Let's say we have a simply supported beam like as shown at 3:45 mark. The beam span is 10 m and the moving load is 50 kN. Is the MAXIMUM shear value at point C (at the midpoint of the beam) be 50 kN x 0.5 = 25 kN (+ and -) and the MAXIMUM moment at the same location be 50 kN x 2.5 = +125 kN-m? Where 2.5 is the ordinate of the moment influence line at point C. Thanks in advance for your reply.

[@DrStructure]

It is important to remember that the shear influence line does not give us any information about the location of max shear in the beam. Rather, it lets us determine the maximum shear at a specific point due to a moving load. The maximum shear in a simply supported beam does not occur at the midpoint of the beam; it takes place close to either support. However, for the simply supported beam you’ve referenced, assuming we want to determine the max shear at the midspan (and not the max shear in the beam), your analysis is correct.

[@gnidnoeled786]

@@DrStructure Your detailed explanation is awesome. It gives me a better understanding of influence line. I really do appreciate it. Salute to you. 👍👏

[@gnidnoeled786]

@@DrStructure With reference to your explanation, the maximum shear is at the support and that's 50 kN x 1 = 50 kN and the maximum moment is at midspan 50 kN x 2.5 = 125 kN-m.

[@DrStructure]

Correct!

[@lifeisbeautiful350]

Thanks so much for this presentation

[@1f7sivakumarponakalla74]

U really nailed it! Thanks a ton!!

[@vuralmert9497]

Awesome.! Greatings from Turkey , Dear Dr.Structure :)

[@rabbiahmed6760]

u explained in the easiest way.tnx a lot

[@bendeng2768]

Very good explanation.

[@obaidurrahman4300]

Good lecture. Thank you ❤❤

[@dayana6532]

This is amazing! Thank you so much for this!

[@saurabhdhadiwal129]

Excellent video

[@DELICKNSUBUGA]

Your the best ❤

[@陳建忠-t6d]

認真的老師.

[@AyaanKhan-ei3gk]

I don't understand how can the pin move vertically up and down ? As you mentioned at 9:15

[@DrStructure]

At 9:15 we place a pair of vertical rollers, not a pin, in the beam. By definition a roller can roll/move. Since they are placed in the vertical directions, the rollers can move/roll up and down. Since we have two of them, they can move independently in opposite directions.

[@AyaanKhan-ei3gk]

@@DrStructure no that's not what I asked , I am asking that there is an internal hinge between two rollers right ? I am asking how can that internal hinge point move up ? Doesn't the pin restrict vertcal motion ?

[@DrStructure]

If you are referring to the real hinge, yes, it can move up/down. A hinge is unlike a pin or a roller. You can view the hinge as a single bolt that connects two beam segments. It allows one segment to rotate relative to the other segment. You can create and experiment with a hinge connection using two balsa wood sticks and a push pin.

[@AyaanKhan-ei3gk]

@@DrStructure hi , thank you so much for your reply , so what I have understood is that a hinge and pin are two separate entities , and point which is hinged on a beam can actually move up and down that's why we have the point moving up , but if we had a pin there then it would have remained at its position and would not have travelled up , right ? Because the pin connection and roller connection resists the vertical movement ,

[@DrStructure]

Correct. The pin and the roller are mechanisms that connect the beam to the support. Neither the pin or the roller can separate itself vertically from the support. The hinge however is not attached to a support, so it’s vertical displacement is not restrained.

[@greatminds8977]

God bless you 🙏

[@Ilsan1111]

thanks for this lecture! awesomeeee

[@ahmetkasmoglu1576]

Great work. İ know a little english but i understand this lesson. Thank you very much.

[@engineersoutlook7]

Excellent tutorial

[@rabin5903]

very convenient explanation thanks...

[@koper1504]

Amazing tutorial, but why don't AD and CD have the same slope? Can you please explain that! minute 13:45 I understand that CD can't rotate because its fixed at B and C. But when BC stays horizontal, the other one should be horizontal as well (AE). Because two lines must remain parallel to each other. Thanks in advance!

[@DrStructure]

The two lines being parallel is a reflection of how shear varies linearly in proportion to the position of the unit load. As long as that hypothesis is true (as long as shear varies linearly in proportion to the load position), then the lines to the left and right of the target point remain parallel, they have the same slope magnitude. However, there are situations in which there is an abrupt change in the shear value where the linearity and proportionality does not hold true. For example, in the beam under this discussion, note that if no load is being applied to segment AD, if all the loads are placed only on segment DC, AD would carries no internal force. In fact, for analysis purposes, we can remove AD altogether and just deal with segment DC. Put it differently, AD carries internal loads only if the load is placed on AD, otherwise no internal shear or moment develops in the segment. So, as the unit load moves across the beam, as soon as it passes point D, the notion of linearity and proportionality does not apply to AD any longer. Meaning, the parallel line hypothesis does not apply here since the linearity and proportionality does not hold true.

[@charrysalce4580]

Very well presented and explained. Thank you so much!

[@DrStructure]

Thank you for the feedback.

[@kitetama13]

thanks for sharing! God bless you

[@saurabhdhadiwal129]

4 yr passed away in college didn't understood the concept but ur video clr my doubt...🙏🙏🙏

[@nisreensalameh1887]

Keep up the good work. Can you do some video series on concrete and steel design? Thank u so much, people like you makes world a much better place

[@sabbirnadim5269]

best tutorial ever

[@gnidnoeled786]

To get the shear influence line at the midspan (at point C) of the simply suppoted beam, a pair of rollers is placed at point C. The left roller is pushing downward and the right is pushing upward. Let's say, I wish to get the shear IL at points A and B by placing a pair of rollers at those points. Is the shear at point A positive, and at B negative based on the pair of rollers placed at points mentioned?

[@DrStructure]

If A and B are points very close to the left and right ends of the beam, then yes, the shear influence line for point A is basically a right triangle with a positive height of almost one (1) at A. The influence line for B is also a right triangle having a height of negative one at B.

[@gnidnoeled786]

@@DrStructure Thanks a lot. I really do understand now how shear IL behaves. I appreciate your continuous replies. 👍👏

[@DrStructure]

You’re welcome!

[@ikbenvoetbal]

Thank you so much. I would bake you a cake. This really helped me a lot. May God bless you and your family

[@Julian-xe3gw]

Just one word: thanks!!

[@hectormunoz4812]

@ 9:17 I am confused as to why the beam EC does not move up as one, but is held down at point C. It is not a pin, so it cannot be held down, can some one explain?

[@DrStructure]

The actual beam does not move up or down at C, the fictitious beam does that. We are inserting an imaginary hinge at C thereby creating a fictitious beam. We then subject that hinge to a shear force causing point C to move up/down in the fictitious beam. Why are we introducing the imaginary hinge and creating a fictitious beam? Because the resulting displaced shape of the fictitious beam corresponds to the shear influence line in the real beam.

[@SHUBHAMKumar-qd1pi]

Awesome video 👍👌👏

[@catobsessed379]

u r awesome dr.!!!!!!!!!!!!!!!! thank you

[@williamsentemtz9652]

Mind blowing!👍👌 5 stars for u

[@sssskhaled4178]

how can I have the answers of the final 3 exercise

[@DrStructure]

Exercise 1: kzbin.info/www/bejne/lX61d4utoaypr6M Exercise 2: kzbin.info/www/bejne/nJS3Za2LZpx-mrM Exercise 3: kzbin.info/www/bejne/iF7WaodrocytqJY

[@asmaulhusna732]

Dr. Structure thank you

[@imbimb1403]

Thank you for such a great explanation. It really helped me .

[@waleedalshammri420]

I’m never see any one make it it simple like this

[@dylandewilzen1864]

How would you go about calculating the influence line if you had one continuous member, with varying cross-section, spanning from A to F as per the above illustration?

[@DrStructure]

If the beam is statically determinate, its cross-section does not impact the shape of the influence line. That is, shear in the beam would not be a function of the cross-sectional properties of the member. but if by continuous you mean indeterminate beam, the qualitative approach would be of little use.

[@feryalighani-missaghi1649]

Great tutorial. On Exercise#1 why you don't rotate the left side beam around B and only shear on the left?

[@DrStructure]

Note the support at the left end of the beam. It looks like a vertical roller, but it behaves a bit differently. The support at A can resist both a horizontal force and a bending moment (similar to a fixed support, but minus a vertical reaction force). This behavior causes a bit of problem for our qualitative approach for drawing the shear influence line. Since the support at A cannot carry any shear (vertical force), no matter where the unit load is placed on the beam, all of the load is going to be counteracted by the vertical reaction at B. That is, the vertical reaction at B is always 1, no matter where the unit load is located. This means, when the unit load is to the right of point C, the left segment of the beam does not carry any vertical (shear) force. That is, as long as the unit load is to the right of C, shear at C (and any other point to the left of C) is zero. Whereas when the unit load is to the left of C, the shear force at C (and any point from C to B) must be equal to 1. This insight yields the rectangular influence line shown in the video.

[@FRSHBL]

Very creative .. Thanks alot

[@xtr3m385]

Shouldn't the maximum shear force at C, VC, be 6 kN? at D on the other hand, VD = 6*1.5 = 9 kN?

[@DrStructure]

Not sure how you arrived at this conclusion, please elaborate.

[@laughinggood]

doc,though the +ve SF convention is right side down and left side up why does the roller moves right side down and left side up which is -ve sign convention for SF????

[@DrStructure]

+Ajay Balan This qualitative approach require that we place a (fictitious) positive shear at the cut point and draw the deformed shape of the beam. This fictitious shear force is not to be confused with the real shear force in the beam. The real shear in the beam is shown by the diagram.

[@FranCoVids]

What if you want to draw the shear influence line on the left and right of a pin/roller support?

[@DrStructure]

The same principle applies. Say, we want to draw the shear influence line for the point just to the right of B for the beam @9:00. Note the influence line for E where the amount of upward movement (of the roller to the right) equals to the amount of the downward movement (of the roller to the left). They each equal 1/2 adding up to 1. Keeping in mind that that total height is always 1 regardless of the position of E, start moving E to the left toward B. As the point gets closer to B, the height above the x axis increases (becomes larger than 1/2, approaching 1 but never reaching it) while the height below the axis gets closer to zero, but it never reaches zero since shear at B is undefined. No matter how small that negative height gets, it results in a downward movement that causes the segment between the hinge at E (now positioned just to the right of B) to rotate clockwise.

[@FranCoVids]

@@DrStructure So, you are basically saying that the shear influence line just to the right of support B will essentially be the same as the shear line for point E? (Just with different heights)

[@DrStructure]

@@FranCoVids Yes!

[@FranCoVids]

@@DrStructure thank you!! You're the best!! 😄

[@muelitassasg]

Thank you for this video!!! do you have the link to the solution to the last 3 questions from this video?

[@DrStructure]

The solutions are provided in the free online course referenced in the video description field.

[@oscarrinn]

I love Dr. Structure!! My structural sicknezz is kured

[@meargbrihanu8139]

thank you so much its very help full , please add some videos on design

[@homebird4765]

@Dr. Structure Please elaborate as to how to draw SF ILD at an intermediate support or simply the ILD for SF at the support of an overhanging beam. This method, although brilliant doesn't seem to work in the said problems.

[@DrStructure]

In beams and frames, there is no shear force at the point of application of a concentrated load, or at the support reactions. That is, at such points shear is undefined. We can talk about shear just to the left (or to the right) of a support, but not at the support. Consider the beam shown @12:42 in this lecture, where the influence line for shear at D is drawn. Note that the overall height of the vertical line at D is one, since the moving load has a magnitude of one. Part of the vertical line is below the x-axis and part of it is above the x-axis. Now, move that vertical line toward B (the roller support). As the line get's closer and closer to B, the lower (below the x-axis) part of it becomes smaller and smaller as the upper part of it (above the x-axis) approaches one. That would be the shear influence line for a point close to an interior support. No matter how close that line gets to B, since it will never reaches B (i.e., shear at B is undefined), there would always be a small part of that vertical line below the x-axis.

[@homebird4765]

@@DrStructure why is shear undefined at the supports? At the point of application of concentrated load mid span , i can understand there are are 2 values of SF at the same point and so SF is undefined there. But why at the support?

[@DrStructure]

Fundamentally, there is no difference between a mid-span point load and a support reaction (also a point load). They are just applied at different locations on the beam. Think of a beam as an (infinite) series of dimensionless points. Now think of the support position as such a point, and the support reaction being a force applied at that point. Let's call the point B, and say, the support reaction at B is 100 kN. For the sum of the forces to vanish, for the force equilibrium to be maintained, therefore, there must be one or two forces (at the two neighboring points) that add up to 100. Let's label the two neighboring points as A and C, and say that there is a force of 50 kN at each of those points so that the sum of all three forces adds up to zero. These balancing forces at A and C are what we call shear forces. By definition, these forces cannot be placed at B. That is, placing the two 50 kN forces at B is equivalent to saying that there is no force (support reaction) being applied at B, since these forces cancel each other out. Whereas the presence of these forces in the neighboring points suggest force propagation through the member, as in the beam is resisting the applied load and consequentially internal (shear) forces develop. That is why there is always a jump in the shear diagram at a support, say, going from -50 kN to +50 kN, with the height of that vertical line being the magnitude of the reaction force at that point.

[@homebird4765]

@@DrStructure understood... Thanks a ton 👍

[@schuyler722]

thank you ma'am. btw where can I find the solution of the practice problem for checking ma'am

[@DrStructure]

The solutions are provided in the (free) course references in the video description field.

[@omegachong2078]

A question Dr., how to draw the shear influence line diagram on a reaction support?

[@DrStructure]

As an internal force, shear is not defined where the beam is subjected to a concentrated (point) load, like at a support. We can talk about the shear force a bit to the right of the support, or a bit to the left of it, but not right at the support. So right at the support, we can only draw the support reaction influence line, not the shear influence line.

[@joseh7612]

how do you draw the influence line if they ask you for a point that's located on the pin

[@DrStructure]

Shear exists, and can be defined, only at points away from pin/roller/fixed supports. Shear force at the pin support is undefined, hence, we cannot draw such an influence line.

[@nasseryariphdpe648]

Great work, what program are you using to make the animation ?

[@DrStructure]

We used Adobe Illustrator and Camtasia Studio for this particular video. And, the talking head was done using CrazyTalk.

[@nasseryariphdpe648]

@@DrStructure Thanks